Recent content by Deanmark

Let $f$ be analytic in the disc $|z| < 1$ and assume that $|f(z)| \le \dfrac{1}{1-|z|}$. Show that $|{f}^{(n)}(0)| ≤ e(n + 1)!$. Any ideas on how to bound $\max|f(z)|$ in the disc?

Isn't $2^0$ the first element and $2^4$ the fifth operation which results in the identity making ${\Bbb{Z}}_{5}$ order 5? I guess this is what the proof is referring to when it claims for x in ${Z}_{d}$ $x^d = 1 $ where d is a prime power.

I am looking at this proof and I am stuck on the logic that $a^{p}$ = 1. For example, consider the group under multiplication without zero, ${Z}_{5}$, wouldn't 2^4 = 1 imply that the order is 4 not 5? We know that if G is a finite abelian group, G is isomorphic to a direct product...

Should we let the complex number u = $\frac{2}{\delta}$ + arg(${w}_{0})$i ?

Consider the function $f(z) = e^{1/z}$, Show that for any complex number ${w}_{0} \ne 0$ and any δ > 0, there exists ${z}_{0} ∈ C$ such that $ 0 < |{z}_{0}| < δ$ and $f({z}_{0}) = {w}_{0}$ I really don't know where to begin on this.

Gotcha. I appreciate it.

A mistake. It's just $2017^{2}$ mod(100) which you can find with a calculator, but if you did not have a calculator could we transform $2017^{2}$ mod(100) further.

$2017^{40} \equiv$ 1 mod (100) $2017^{82} = 2017^{80} \cdot 2017^{2}$ $ 2017^{2} - 1 \equiv 0 \ mod (100)$ 88??

I don't know anything about it. We only went of Fermat's and Euler's generalization of ax$\equiv$ b mod(m). The class is a Ring Theory class so we aren't really diving that deeply into the number theory.

What are the last two digits of ${2017}^{82}$? We have only learned Fermat's little theorem and Euler's generalization so I am not sure how they apply?

The parts of my wall that have yet to be punched thank you.

Compute the remainder of 2^(2^17) + 1 when divided by 19. The book says to first compute 2^17 mod 18 but I don’t understand why we go to mod 18. Advice would be appreciated

Let f(z) = $e^{e^{z}}$ . Find Re(f) and Im(f). I don't know how to deal with the exponential within an exponential. Does anybody know how to deal with this?

I think that is the best way to go. If we (I) can show it converges to the same constant, then the problem is done.

The question is, how can we prove $\lim_{{n}\to{n+1}}\int_{n}^{n+1}cos(2x^2)/2 \,dx = 0$? The best I have been able to come up with is the fact that $\lim_{{n}\to{n+1}}\ \int_{0}^{n+1}cos(2x^2)/2 \,dx - \int_{0}^{n}cos(2x^2)/2 \,dx$ = $\int_{n}^{n+1}cos(2x^2)/2 \,dx$ and each function is...